Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{6z}{8(z + 2)} \times \dfrac{5z + 10}{-5} $
When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ 6z \times (5z + 10) } { 8(z + 2) \times -5 } $ $ a = \dfrac {6z \times 5(z + 2)} {-5 \times 8(z + 2)} $ $ a = \dfrac{30z(z + 2)}{-40(z + 2)} $ We can cancel the $z + 2$ so long as $z + 2 \neq 0$ Therefore $z \neq -2$ $a = \dfrac{30z \cancel{(z + 2})}{-40 \cancel{(z + 2)}} = -\dfrac{30z}{40} = -\dfrac{3z}{4} $